binomial random variable applications, conditional lect4a
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4. Binomial Random Variable Approximations,
Conditional Probability Density Functions
and Stirlings Formula
LetX represent a Binomial r.v as in (3-42). Then from (2-30)
Since the binomial coefficient grows quiterapidly with n, it is difficult to compute (4-1) for large n. In
this context, two approximations are extremely useful.
4.1 The Normal Approximation (Demoivre-Laplace
Theorem) Suppose withp held fixed. Then for kinthe neighborhood of np, we can approximate
( ) =
=
==
2
1
2
1
.)(21
k
kk
knkk
kk
n qpk
nkPkXkP (4-1)
!)!(!
kknn
k
n
=
nnpq
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(4-2).2
1 2/)( 2 npqnpkknk enpq
qpk
n
Thus if and in (4-1) are within or around theneighborhood of the interval we can
approximate the summation in (4-1) by an integration. In
that case (4-1) reduces to
where
We can express (4-3) in terms of the normalized integral
that has been tabulated extensively (See Table 4.1).
1k 2k
,, npqnpnpqnp +
( ) ,2
1
2
1 2/
2/)(
21
22
1
22
1
dyedxenpq
kXkP yx
x
npqnpxk
k
== (4-3)
)(
2
1)(
0
2/2 xerfdyexerfx
y ==
(4-4)
., 221
1npq
npkx
npq
npkx
=
=
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For example, if and are both positive ,we obtain
Example 4.1: A fair coin is tossed 5,000 times. Find the
probability that the number of heads is between 2,475 to
2,525.
Solution: We need Here n is large so
that we can use the normal approximation. In this caseso that and Since
and the approximation is valid for
and Thus
Here
( ) ).()( 1221 xerfxerfkXkP =
1x 2x
).525,2475,2( XP
(4-5)
,2
1=p
500,2=np .35npq ,465,2= npqnp
,535,2=+ npqnp 475,21=k
.525,22 =k
( ) =2
1
2
.2
1 2/21
x
x
y dyekXkP
.75,
75 2211 ==== npq
npkxnpq
npkx
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2
1)(
2
1)(erf
0
2/2 == xGdyexx
y
x erf(x) x erf(x) x erf(x) x erf(x)
0.05
0.100.150.20
0.25
0.30
0.350.400.45
0.50
0.55
0.600.65
0.700.75
0.01994
0.039830.059620.07926
0.09871
0.11791
0.136830.155420.17364
0.19146
0.20884
0.225750.24215
0.258040.27337
0.80
0.850.900.95
1.00
1.05
1.101.151.20
1.25
1.30
1.351.40
1.451.50
0.28814
0.302340.315940.32894
0.34134
0.35314
0.364330.374930.38493
0.39435
0.40320
0.411490.41924
0.426470.43319
1.55
1.601.651.70
1.75
1.80
1.851.901.95
2.00
2.05
2.102.15
2.202.25
0.43943
0.445200.450530.45543
0.45994
0.46407
0.467840.471280.47441
0.47726
0.47982
0.482140.48422
0.486100.48778
2.30
2.352.402.45
2.50
2.55
2.602.652.70
2.75
2.80
2.852.90
2.953.00
0.48928
0.490610.491800.49286
0.49379
0.49461
0.495340.49597
0.49653
0.49702
0.49744
0.497810.49813
0.498410.49865
Table 4.1
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Since from Fig. 4.1(b), the above probability is given
by
where we have used Table 4.1
4.2. The Poisson ApproximationAs we have mentioned earlier, for large n, the Gaussian
approximation of a binomial r.v is valid only ifp is fixed,
i.e., only if and what if np is small, or if itdoes not increase with n?
,01>np .1>>npq
( ).258.0)7.0(erf =
Fig. 4.1
x
(a)
1x 2x
2/2
2
1 xe
0,0 21 >> xx
x
(b)
1x 2x
2/2
2
1 xe
0,0 21 >< xx
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Obviously that is the case if, for example, as
such that is a fixed number.
Many random phenomena in nature in fact follow this
pattern. Total number of calls on a telephone line, claims in
an insurance company etc. tend to follow this type of
behavior. Consider random arrivals such as telephone calls
over a line. Let n represent the total number of calls in the
interval From our experience, as we haveso that we may assume Consider a small interval of
duration as in Fig. 4.2. If there is only a single call
coming in, the probabilityp of that single call occurring inthat interval must depend on its relative size with respect to
T.
0p ,n=np
T).,0( T n.Tn =
1 2 n
0 T
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Hence we may assume Note that as
However in this case is a constant,
and the normal approximation is invalid here.
Suppose the interval in Fig. 4.2 is of interest to us. A call
inside that interval is a success (H), whereas one outside isa failure (T ). This is equivalent to the coin tossing
situation, and hence the probability of obtaining k
calls (in any order) in an interval of duration is given bythe binomial p.m.f. Thus
and here as such that It is easy to
obtain an excellent approximation to (4-6) in that situation.
To see this, rewrite (4-6) as
.T
p = 0p .T
==
=
T
Tnp
)( kPn
,)1(!)!(
!)( knkn pp
kkn
nkP
= (4-6)
0, pn .=np
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.)/1()/1(
!112111
)/1(!
)()1()1()(
k
nk
knk
kn
nn
knk
nn
nnpk
np
n
knnnkP
=
+
=
(4-7)
Thus,
!
)(lim,0,
=
= ek
kPk
nnppn
(4-8)
since the finite products as well
as tend to unity as and
The right side of (4-8) represents the Poisson p.m.f and the
Poisson approximation to the binomial r.v is valid in
situations where the binomial r.v parameters n andp
diverge to two extremes such that theirproduct np is a constant.
n
k
nn
11
21
11
k
n
1 ,n
.1lim
=
e
n
n
n
)0,( pn
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Example 4.2: Winning a Lottery: Suppose two million
lottery tickets are issued with 100 winning tickets among
them. (a) If a person purchases 100 tickets, what is theprobability of winning? (b) How many tickets should one
buy to be 95% confident of having a winning ticket?
Solution: The probability of buying a winning ticket
Here and the number of winning ticketsXin the n
purchased tickets has an approximate Poisson distribution
with parameter Thus
and (a) Probability of winning
.105102
100
ticketsofno.Total
ticketswinningofNo. 56
=
==p
,100=n
.005.0105100
5
===
np,
!)(
kekXP
k==
.005.01)0(1)1( ====
eXPXP
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(b) In this case we need
But or Thus one needs to
buy about 60,000 tickets to be 95% confident of having a
winning ticket!
Example 4.3: A space craft has 100,000 components
The probability of any one component being defectiveis The mission will be in danger if five or
more components become defective. Find the probability of
such an event.Solution: Here n is large andp is small, and hence Poisson
approximation is valid. Thus
and the desired probability is given by
.95.0)1( XP
.320lnimplies95.01)1( == eXP
3105 5 == nnp .000,60n
( )n
).0(102 5 p
,2102000,100 5 === np
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{ } ( ){ }
.)(
)(|)()|(
BP
BxXPBxXPBxFX
==
Thus the definition of the conditional distribution depends
on conditional probability, and since it obeys all probability
axioms, it follows that the conditional distribution has the
same properties as any distribution function. In particular
Further
( ){ }
( ){ }.0
)(
)(
)(
)()|(
,1
)(
)(
)(
)()|(
==
=
==+
=+
BP
P
BP
BXPBF
BP
BP
BP
BXPBF
X
X
(4-12)
(4-11)
( ){ }
),|()|(
)(
)()|)((
12
2121
BxFBxF
BP
BxXxPBxXxP
XX
=
+ + > +
(4-49)
1112 ( 1)2
(1 )
! 2 .n n
nn
n n e + +
>
1 11 1
2 ! 2n nn n+ + (4 50)
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Stirlings formula also follows from the asymptoticexpansion for the Gamma function given by
Together with the above expansion can
be used to compute numerical values for realx.
For a derivation of (4-51), one may look into Chapter 2
of the classic text by Whittaker and Watson (Modern
Analysis).
We can use Stirlings formula to obtain yet anotherapproximation to the binomial probability mass
( 1) ( ),x x x + =
12 1 122 22 ! 2 .n nn nn nn e e n n e e ++ + < < (4-50)
(4-51){ }
12
2 3 4
1391 1 112 286 51840( ) 2 1 ( )
xx
x x x xx e x o
= + + +
function. Since
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using (4-50) on the right side of (4-52) we obtain
and
where
and
Notice that the constants c1 and c2 are quite closeto each other.
( ) ( )1 2 ( )k n kk n k np nqn
n k k k n k
n
p q ck
>
!,
( )! !
k n k k n k n n
p q p q
n k kk
=
(4-52)
{ }1 1 112 1 12( ) 121
n n k k
c e +
={ }1 1 112 12( ) 1 12 1
2 .n n k k c e + +
=
( ) ( )2 2 ( )k n kk n k np nq
nn k k k n k
n
p q ck